A finite simple group with subgroup for every divisor of $|G|$ is abelian

Question

Let $G$ be a finite simple group.

Assume that for every positive integer $d$ that divides $|G$, there is a subgroup $H$ of $G$ such that $|H| = d$.

Prove that $G$ is abelian.

So I'm really out of ideas here...

I don't know much theorems that is about both simple and abelian groups...

I was thinking about - since $G$ is simple, and $Z(G)$ is always a normal subgroup, $Z(G)$ can be either $G$, or $\{e\}$.

So we only need to prove that $Z(G)$ is not trivial, and we are done..

Any hints on how to do that?

score 7 · Accepted Answer · answered Jun 19 '18 at 04:43

7

Let $G$ be a group of order $n$, and let $p$ be the least prime factor of $n$. There is a theorem to the effect that if $G$ has a subgroup $H$ of index $p$, then $H$ is normal in $G$. (Hint: consider the action of $G$ on the left cosets of $H$). If $G$ is simple, then $H$ must be trivial, and so $G$ has order $p$.

answered Jun 19 '18 at 04:43

Angina Seng

158,341

See: https://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index. (These type of arguments are not obvious unless you've seen them before.) – pancini Jun 19 '18 at 05:01
And then, it is isomorphic to a cyclic group and therefore abelian? – ChikChak Jun 19 '18 at 05:04
@ChikChak Exactly! – Angina Seng Jun 19 '18 at 05:10
Thank you, this is a really great solution. And yet, out of curiosity, could there be any way doing it as I started?(trying to prove that $|G|=Z(G)$)? – ChikChak Jun 19 '18 at 05:14

A finite simple group with subgroup for every divisor of $|G|$ is abelian

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